SOLUTION: 0 = 5(x^3) + 10 + 10(x^2) solve for x? or factor?

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Question 29300: 0 = 5(x^3) + 10 + 10(x^2)
solve for x? or factor?
Answer by longjonsilver(2297) About Me  (Show Source):
You can put this solution on YOUR website!
Not too sure how to approach this apart from a numerical one... We have x%5E3%2B2x%5E2%2B2=0. This does not factorise easily.

Approach is:
pick value of x
find the value of y
pick another value of x
when sign of y changes, we have "straddled" the solution, the root, the answer.
we then look at a more precise value of x and repeat the process.

So,
start with x=-1... y=%28-1%29%5E3+%2B+2%28-1%29%5E2+%2B+2 --> -1 + 2 + 2 --> +ve
now, try x = -2... y=%28-2%29%5E3+%2B+2%28-2%29%5E2+%2B+2 --> -8 + 8 + 2 --> +ve but getting closer to zero
now try x = -3... y=%28-3%29%5E3+%2B+2%28-3%29%5E2+%2B+2 --> -27 + 18 + 2 --> -ve

So the root lies between x=-2 and x=-3.

Now pick x=-2.5... y=%28-2.5%29%5E3+%2B+2%28-2.5%29%5E2+%2B+2 --> -ve, so the root lies between x=-2 and x=-2.5

So, pick x=-2.4... y=%28-2.4%29%5E3+%2B+2%28-2.4%29%5E2+%2B+2 --> -ve, Just. The root lies between x=-2 and x=-2.4

Pick x=-2.3... y=%28-2.3%29%5E3+%2B+2%28-2.3%29%5E2+%2B+2 --> +ve, so the root lies now between x=-2.3 and x=-2.4.

So now you try say x=-2.35 and work out the bounds of the root etc.

Looking at the graph, we do indeed see a solution between -2 and -3.

graph%28300%2C300%2C-5%2C5%2C-10%2C10%2Cx%5E3%2B2x%5E2%2B2%29

graph%28300%2C300%2C-4%2C-2%2C-5%2C5%2Cx%5E3%2B2x%5E2%2B2%29 is a closer look at the root.
jon